This question about finite posets is a special case of my research problem about filters (because for finite posets the set of filters is (dually) isomorphic to the set of elements).
Consider a (finite) lattice $\mathfrak{A}$ with least element. I denote binary meet, binary join, least, and greatest elements correspondingly as $\sqcap$, $\sqcup$, $\bot$, $\top$.
I call star of an element $a$ the set $$\star a = \{ x\in\mathfrak{A} \mid x\sqcap a\ne\bot \}.$$
It is easy to show that for boolean lattices $\star x = \{ \overline X \mid X\in\mathfrak{A},X\not\geq x\}$. Thus (as this is a lattice isomorphism) for boolean lattices we have $$\forall a,b\in\mathfrak{A}: (\star a=\star b \Rightarrow a=b).$$
Can the last formula hold for non-boolean (finite) (distributive?) lattices? What are necessary and/or sufficient conditions for our lattice for this formula to hold?
This is not true for a general finite, distributive lattice. Indeed, consider the lattice which has 6 elements: $\perp,\top,x,y,x\sqcap y, x\sqcup y$, which is distributive. Notice that $\star x=\{\top,x,y,x\sqcup y,x\sqcap y\}=\star y$.
Edit: Here's a proof that if $L$ is a finite, distributive lattice with $\star x=\star y\implies x=y$, then $L$ must be boolean. Instead of working with $\star$, let's work with the complement, call it $\circ x=L\setminus\star x=\{y\in L:y\sqcap x=\perp\}$. I first want to claim that $\circ x$ has a largest element. Indeed, this follows from distributivity: if $y,z\in\circ x$, then $x\sqcap(y\sqcup z)=(x\sqcap y)\sqcup(x\sqcap z)=\perp\sqcup\perp=\perp$, so $y\sqcup z\in\circ x$. Hence, since $L$ is finite, we can define $x^\circ=\max(\circ x)$. Now, notice that $\circ x$ is a down-set in the poset induced by the meet and join, so we have $\circ x=\circ y\iff x^\circ=y^\circ$. We need one extra result which follows from definition chasing, so I'll omit the proof: $(x\sqcap y)^\circ=x^\circ\sqcup y^\circ$.
Now, We know that $\top=\perp^\circ=(x\sqcap x^\circ)^\circ=x^\circ\sqcup x^{\circ\circ}$. But, of course, by definition $x^\circ\sqcap x^{\circ\circ}=\perp$. Thus, if $y=x^\circ$ for some $x\in L$, we have $y\sqcap y^\circ=\perp$ and $y\sqcup y^\circ=\top$; in other words, all of these elements have complements. Finally, remember that we assumed $x^\circ=y^\circ\implies x=y$, so actually $f:L\to L:x\mapsto x^\circ$ is an injection. But since $L$ is finite, $f$ must actually be a bijection, so every element $y\in L$ is of the form $y=x^\circ$ for some $x\in L$. In other words, all elements of $L$ have complements, so since $L$ is distributive, it must be boolean.
As a final thought, it is worth pointing out that non-distributive lattices can satisfy your formula. For instance, consider the lattice with elements $\perp,\top,x,y,z$ where $\{x,y,z\}$ is an antichain. Certainly this lattice isn't distributive. On the other hand, $\star\top=\{x,y,z,\top\}$, $\star x=\{x,\top\}$, $\star y=\{y,\top\}$ and $\star z=\{z,\top\}$.